What is laplacetransform 3\delta(t)+2+4e^{-3t+2} ?

The answer to laplacetransform 3\delta(t)+2+4e^{-3t+2} is 3+2/s+(4e^2)/(s+3)

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laplacetransform 3\delta(t)+2+4e^{-3t+2}

laplace transform

3 δ (t) + 2 + 4 e^{- 3 t + 2}

3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

Solution steps

L {3 δ (t) + 2 + 4 e^{- 3 t + 2}}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 3 L {δ (t)} + L {2} + 4 e^{2} L {e^{- 3 t}}

L {δ (t)} : 1

L {2} : \frac{2}{s}

L {e^{- 3 t}} : \frac{1}{s + 3}

= 3 \cdot 1 + \frac{2}{s} + 4 e^{2} \frac{1}{s + 3}

Refine

31 + \frac{2}{s} + 4 e^{2} \frac{1}{s + 3} : 3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

= 3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

\frac{\partial}{\partial x} (\frac{y}{1 + x ^{2}})

(60 + 2 t)^{\frac{2}{3}}

f (x) = 3^{4 x + 5}

y = - 1 - 8 x^{2}, (- 2, - 33)

f (x) = x^{\frac{2}{5}}

What is laplacetransform 3\delta(t)+2+4e^{-3t+2} ?
The answer to laplacetransform 3\delta(t)+2+4e^{-3t+2} is 3+2/s+(4e^2)/(s+3)